I came to this problem when doing the exercise that the polydisc $\Delta(0,1)^n=\prod\limits_{n}\Delta(0,1)$ in $\mathbb{C}^n$ is not biholomorphic to the unit ball $\mathbb{B}^n$ in $\mathbb{C}^n$.We can show that the group of automorphisms of $\Delta(0,1)^n$ preserving the origin is the semidirect of $U^n(1)=\prod\limits_{n}U(1)$ and the permutation group $S_n$(the semidirect product here means that $U^n(1)$ si a normal subgroup of $U^n(1)\rtimes S_n$), while that of $\mathbb{B}^n$ is $U(n)$.
Then I do not know how to show that the above two groups are not isomorphic.Intuitively, topologically the (real) dimension of $U^n(1)\rtimes S_n$ is $n$ , while that of $U(n)$ is $\frac{n(n-1)}{2}$ (is it right?), so they are not isomrphic.
Clearly it is not a rigorous proof.
Will someone be kind enough to give me some hints for this problem?Thank you very much!
Best Answer
One simple enough way is to count the square roots of the identity. The $n=2$ case is illustrative. The equation $((z_1,z_2),\sigma)^2 = ((z_1^2,z_2^2),\sigma^2) = ((1,1),e)$ gives us 8 square roots of the identity in $U^2(1)\rtimes S_2$, while it is easily verified that there are only 6 square roots of the identity matrix in $U(2)$. Generalizing to all $n>1$ shouldn't be too hard.
Alternatively, you could set up a map $U^n(1)\rightarrow U(n)$ and look at where tuples with all but one entry equal to $1$ are sent, then verify that this cannot be a normal subgroup. This may require more creativity (I haven't worked it out), but would also be prettier.